Question

Without actually solving the simultaneous equations given below, decide which simultaneous equations have unique solutions, no solution or infinitely many solutions.

$\left(\mathrm{vi}\right)\frac{x}{2}+\frac{y}{3}=4;\frac{x}{4}+\frac{y}{6}=2$

Answer 1

$$\begin{gathered} \frac{x}{2}+\frac{y}{3}=4 \\ \frac{x}{4}+\frac{y}{6}=2 \end{gathered}$$

Comparing the above equations with the equations $a_{1}x+b_{1}y=c_1\mathrm{and}{a}_{2}x+b_{2}y=c_2,\mathrm{we}\mathrm{get}$:

$$\begin{gathered} a_1=\frac{1}{2}{b}_1=\frac{1}{3}{c}_1=4 \\ {a}_2=\frac{1}{4}{b}_2=\frac{1}{6}{c}_2=2 \\ \therefore \frac{a_1}{a_2}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=2;\frac{b_1}{b_2}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}=2\mathrm{and}\frac{c_1}{c_2}=\frac{4}{2}=2 \end{gathered}$$

$$\begin{gathered} \mathrm{i}.\mathrm{e}.,\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ \therefore \mathrm{The}\mathrm{given}\mathrm{equations}\mathrm{have}\mathrm{infinitely}\mathrm{many}\mathrm{solutions}. \end{gathered}$$